Trigonometric functions: Values of the given expressions

To find the values of the given expressions, we can use the properties of inverse trigonometric functions.

1. cos⁡−1(cos⁡(136))\cos^{-1}(\cos(\frac{13}{6}))

The cosine function is periodic and has a range of [0,π][0, \pi] for its inverse. Thus, we need to find an angle that is equivalent to 136\frac{13}{6} within the range of [0,π][0, \pi]:

1. First, convert 136\frac{13}{6} into degrees: $ \frac{13}{6} \times \frac{180}{\pi} \approx 123.69^\circ $
2. Now, find the reference angle: $ 123.69^\circ - 180^\circ = -56.31^\circ $
3. The cosine is positive in the second quadrant, thus: $ \cos^{-1}(\cos(\frac{13}{6})) = \pi - (\frac{13}{6} - 2\pi k) \text{ for some integer } k $

Since k=0k = 0: $ = \pi - \frac{13}{6} = \frac{6\pi - 13}{6} $ So, $ \cos^{-1}(\cos(\frac{13}{6})) = \frac{6\pi - 13}{6} $

2. tan⁡−1(tan⁡(76))\tan^{-1}(\tan(\frac{7}{6}))

The tangent function is periodic with a period of π\pi and has a range of (−π2,π2)(- \frac{\pi}{2}, \frac{\pi}{2}). Therefore, we need to find an equivalent angle for 76\frac{7}{6}:

1. Convert 76\frac{7}{6} into degrees: $ \frac{7}{6} \times \frac{180}{\pi} \approx 66.67^\circ $
2. Since 66.67∘66.67^\circ is in the first quadrant, we need to adjust it to fall within the range of the inverse tangent:
   • The equivalent angle can be obtained by subtracting π\pi: $ = \frac{7}{6} - \pi = \frac{7}{6} - \frac{6}{6}\pi = -\frac{\pi}{6} $

So, $ \tan^{-1}(\tan(\frac{7}{6})) = -\frac{\pi}{6} $

Conclusion

1. cos⁡−1(cos⁡(136))=6π−136\cos^{-1}(\cos(\frac{13}{6})) = \frac{6\pi - 13}{6}
2. tan⁡−1(tan⁡(76))=−π6\tan^{-1}(\tan(\frac{7}{6})) = -\frac{\pi}{6}

Sample Answer